| Content On This Page | ||
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| Division Algorithm for Polynomials | Remainder Theorem | Factor Theorem |
| Division of Polynomial by Polynomial | ||
Polynomial Theorems and Division
Division Algorithm for Polynomials
Polynomial Analogue of Euclid's Division Lemma
In arithmetic, we learn about the division process for integers. For example, when we divide $17$ by $5$, we get a quotient of $3$ and a remainder of $2$. We can write this relationship as $17 = 5 \times 3 + 2$. This concept is formally stated as Euclid's Division Lemma (or the Division Algorithm for Integers).
A similar, fundamental theorem exists for polynomials, providing a formal structure for polynomial division. The Division Algorithm for Polynomials states that whenever we divide a polynomial by another non-zero polynomial, the result can be expressed in terms of a unique quotient polynomial and a unique remainder polynomial, with specific conditions on the remainder.
Statement of the Division Algorithm for Polynomials
Statement: Given any two polynomials, $P(x)$ and $G(x)$, with $G(x)$ being a non-zero polynomial, it is always possible to find unique polynomials $Q(x)$ (the quotient) and $R(x)$ (the remainder) such that the following relationship holds:
$P(x) = G(x) \cdot Q(x) + R(x) $
[Division Algorithm for Polynomials]
This equation must satisfy the condition that either the remainder polynomial $R(x)$ is the zero polynomial (i.e., $R(x) = 0$) or the degree of $R(x)$ is strictly less than the degree of the divisor polynomial $G(x)$.
Let's define the parts of this statement:
- $P(x)$: This is the dividend, the polynomial that is being divided.
- $G(x)$: This is the divisor, the non-zero polynomial by which we are dividing $P(x)$. $G(x)$ must not be the zero polynomial.
- $Q(x)$: This is the quotient, the polynomial resulting from the division.
- $R(x)$: This is the remainder, the polynomial left over after the division.
The condition on $R(x)$ is critical for ensuring the uniqueness of the quotient and remainder, and for the process of division to terminate:
- If $R(x) = 0$: This means that $P(x)$ is perfectly divisible by $G(x)$. In this case, $G(x)$ is a factor of $P(x)$, and $P(x) = G(x) \cdot Q(x)$.
- If $R(x) \neq 0$: The degree of $R(x)$ must be strictly less than the degree of $G(x)$. For example, if you are dividing by a polynomial of degree 2 (like $x^2+1$), the remainder must be a polynomial of degree 1 (like $ax+b$) or degree 0 (a constant), or the zero polynomial. It cannot be degree 2 or higher. This condition guarantees that the division process terminates.
Analogy with Integer Division
The Division Algorithm for Polynomials is a direct parallel to the Division Algorithm for Integers (Euclid's Division Lemma). Let $a$ be an integer (dividend) and $b$ be a non-zero integer (divisor). The integer algorithm states that there exist unique integers $q$ (quotient) and $r$ (remainder) such that:
$$ a = bq + r $$where the remainder $r$ satisfies the condition $0 \le r < |b|$.
Comparing the two algorithms:
| Concept | Integer Division ($a = bq + r$) | Polynomial Division ($P(x) = G(x) Q(x) + R(x)$) |
|---|---|---|
| Dividend | $a$ (an integer) | $P(x)$ (a polynomial) |
| Divisor | $b$ (a non-zero integer) | $G(x)$ (a non-zero polynomial) |
| Quotient | $q$ (a unique integer) | $Q(x)$ (a unique polynomial) |
| Remainder | $r$ (a unique integer) | $R(x)$ (a unique polynomial) |
| Remainder Condition | $0 \le r < |b|$ (remainder is non-negative and strictly less than the absolute value of the divisor) | $R(x) = 0$ or degree$(R(x)) <$ degree$(G(x))$ (remainder is zero or its degree is strictly less than the degree of the divisor) |
| Exact Division / Factor | If $r=0$, $b$ is a factor of $a$. | If $R(x)=0$, $G(x)$ is a factor of $P(x)$. |
The absolute value condition $|b|$ for integers is replaced by the degree condition for polynomials. The remainder being non-negative for integers has no direct polynomial analogue in terms of value, but the zero remainder indicates exact divisibility in both cases.
Examples Illustrating the Division Algorithm
Let's look at a couple of examples to see the Division Algorithm in action. The actual process of performing polynomial division to find $Q(x)$ and $R(x)$ is detailed in a later section (Division of Polynomial by Polynomial). Here, we will state the results of the division and verify that they satisfy the algorithm.
Example 1. Let $P(x) = x^2 + 3x + 2$ and $G(x) = x + 1$. Divide $P(x)$ by $G(x)$ and express the result in the form of the Division Algorithm.
Answer:
We are dividing the polynomial $P(x) = x^2 + 3x + 2$ by the polynomial $G(x) = x + 1$.
Performing the polynomial division (we'll learn how to do this later), we find that:
Quotient $Q(x) = x + 2$
Remainder $R(x) = 0$
Now, let's verify if these unique $Q(x)$ and $R(x)$ satisfy the Division Algorithm equation: $P(x) = G(x) \cdot Q(x) + R(x)$.
Substitute the polynomials:
$$ x^2 + 3x + 2 = (x + 1)(x + 2) + 0 $$Let's expand the right side:
$$ (x + 1)(x + 2) + 0 = (x \times x + x \times 2 + 1 \times x + 1 \times 2) + 0 $$ $$ = (x^2 + 2x + x + 2) + 0 $$ $$ = x^2 + 3x + 2 $$So, the equation holds: $x^2 + 3x + 2 = x^2 + 3x + 2$.
Next, check the remainder condition: The remainder is $R(x) = 0$. The condition states that either $R(x)=0$ or degree$(R(x)) <$ degree$(G(x))$. Since $R(x)=0$, the first part of the condition is met.
In this case, since the remainder is 0, the divisor $G(x) = x+1$ is a factor of the dividend $P(x) = x^2+3x+2$. Indeed, $x^2+3x+2$ can be factored as $(x+1)(x+2)$.
Example 2. Let $P(x) = 3x^3 + x^2 + 2x + 5$ and $G(x) = x^2 + 2x + 1$. Divide $P(x)$ by $G(x)$ and express the result in the form of the Division Algorithm.
Answer:
We are dividing $P(x) = 3x^3 + x^2 + 2x + 5$ by $G(x) = x^2 + 2x + 1$.
Performing the polynomial division, we find that:
Quotient $Q(x) = 3x - 5$
Remainder $R(x) = 9x + 10$
Now, let's verify if these unique $Q(x)$ and $R(x)$ satisfy the Division Algorithm equation: $P(x) = G(x) \cdot Q(x) + R(x)$.
Substitute the polynomials:
$$ 3x^3 + x^2 + 2x + 5 = (x^2 + 2x + 1)(3x - 5) + (9x + 10) $$Let's expand the product $(x^2 + 2x + 1)(3x - 5)$:
$$ (x^2 + 2x + 1)(3x - 5) = x^2(3x - 5) + 2x(3x - 5) + 1(3x - 5) $$ $$ = (x^2 \times 3x - x^2 \times 5) + (2x \times 3x - 2x \times 5) + (1 \times 3x - 1 \times 5) $$ $$ = (3x^3 - 5x^2) + (6x^2 - 10x) + (3x - 5) $$Combine like terms in the expanded product:
$$ = 3x^3 + (-5x^2 + 6x^2) + (-10x + 3x) - 5 $$ $$ = 3x^3 + x^2 - 7x - 5 $$Now, add the remainder $R(x) = 9x + 10$ to this product:
$$ (3x^3 + x^2 - 7x - 5) + (9x + 10) $$Combine like terms:
$$ = 3x^3 + x^2 + (-7x + 9x) + (-5 + 10) $$ $$ = 3x^3 + x^2 + 2x + 5 $$This resulting polynomial $3x^3 + x^2 + 2x + 5$ matches the original dividend $P(x)$. The equation $P(x) = G(x) \cdot Q(x) + R(x)$ holds.
Next, check the remainder condition: The degree of the remainder $R(x) = 9x + 10$ is $1$. The degree of the divisor $G(x) = x^2 + 2x + 1$ is $2$. Since $1 < 2$, the condition degree$(R(x)) <$ degree$(G(x))$ is satisfied.
The Division Algorithm for Polynomials is a fundamental theorem that guarantees that for any polynomial division (by a non-zero polynomial), there exists a unique result in the form of a quotient and a remainder that satisfy the stated conditions. This provides the theoretical basis for the process of polynomial long division and is essential for proving other important theorems like the Remainder Theorem and the Factor Theorem.
Remainder Theorem
Connecting Division and Evaluation
The Remainder Theorem is a powerful shortcut in polynomial algebra that simplifies the process of finding the remainder when a polynomial is divided by a simple linear polynomial. Instead of performing the potentially lengthy process of polynomial long division, the theorem states that the remainder can be found by simply evaluating the polynomial at a specific value.
This theorem builds upon the Division Algorithm for Polynomials and provides a direct link between polynomial division and the concept of evaluating a polynomial at a particular number.
Statement of the Remainder Theorem
Statement: Let $P(x)$ be any polynomial of degree greater than or equal to one. When $P(x)$ is divided by the linear polynomial $(x - c)$, the remainder is equal to the value of the polynomial evaluated at $x = c$, i.e., the remainder is $P(c)$.
Let's clarify the components:
- $P(x)$: This is the dividend polynomial. It can be any polynomial with a degree of 1 or higher.
- $(x - c)$: This is the divisor, a linear polynomial where $c$ is a constant number. The form $(x-c)$ is specific; if the divisor is like $(x+c)$, it should be thought of as $(x - (-c))$, so the value of $c$ for the theorem is $-c$. The number $c$ is the zero of the linear divisor $(x-c)$, because setting $x-c=0$ gives $x=c$.
- $P(c)$: This is the value of the polynomial $P(x)$ obtained by substituting $c$ for every instance of the variable $x$.
The theorem states that the result of evaluating the polynomial $P(x)$ at the number $c$ is precisely the number that would be left over as the remainder if you were to divide $P(x)$ by $(x-c)$ using long division.
Proof of the Remainder Theorem
The Remainder Theorem is a direct consequence of the Division Algorithm for Polynomials.
According to the Division Algorithm, when a polynomial $P(x)$ is divided by a non-zero polynomial $G(x)$, there exist unique polynomials $Q(x)$ (quotient) and $R(x)$ (remainder) such that:
$$ P(x) = G(x) \cdot Q(x) + R(x) $$where either $R(x) = 0$ or the degree of $R(x)$ is strictly less than the degree of $G(x)$.
In the case of the Remainder Theorem, the divisor $G(x)$ is the linear polynomial $(x - c)$. The degree of $(x - c)$ is $1$.
Applying the Division Algorithm with $G(x) = (x-c)$, we get:
$P(x) = (x - c) \cdot Q(x) + R(x)$
... (i) [From Division Algorithm]
The condition on the remainder $R(x)$ is that its degree must be strictly less than the degree of the divisor $(x-c)$. Since the degree of $(x-c)$ is $1$, the degree of $R(x)$ must be less than $1$. The only polynomials with degree less than $1$ are constant polynomials (degree 0) or the zero polynomial. This means $R(x)$ must be a constant value.
Let this constant remainder be $k$. So, $R(x) = k$.
Substitute $R(x) = k$ into equation (i):
$$ P(x) = (x - c) Q(x) + k $$This equation is an identity, meaning it holds true for all real values of $x$. Let's choose a specific value for $x$: let $x = c$. Substitute $x = c$ into the equation:
$$ P(c) = (c - c) Q(c) + k $$Since $c - c = 0$, the term $(c - c) Q(c)$ becomes $0 \times Q(c) = 0$.
$$ P(c) = 0 + k $$ $$ P(c) = k $$Since we established that the remainder $R(x)$ is the constant $k$, and we found that $k = P(c)$, we can conclude that the remainder $R(x)$ is equal to $P(c)$.
Thus, the Remainder Theorem is proven: when a polynomial $P(x)$ is divided by $(x-c)$, the remainder is $P(c)$.
Examples Using the Remainder Theorem
The Remainder Theorem provides a quick way to find the remainder without performing polynomial long division, which is particularly useful when the divisor is linear.
Example 1. Find the remainder when the polynomial $P(x) = x^3 - 3x^2 + 2x - 1$ is divided by $(x - 2)$.
Answer:
The polynomial is $P(x) = x^3 - 3x^2 + 2x - 1$.
The divisor is the linear polynomial $(x - 2)$. This is in the form $(x - c)$, where $c = 2$.
According to the Remainder Theorem, the remainder when $P(x)$ is divided by $(x - 2)$ is equal to $P(c)$, which in this case is $P(2)$.
We need to evaluate $P(x)$ at $x = 2$. Substitute $x = 2$ into the expression for $P(x)$:
$$ P(2) = (2)^3 - 3(2)^2 + 2(2) - 1 $$Now, calculate the value following the order of operations:
$$ P(2) = 8 - 3(4) + 4 - 1 $$ $$ P(2) = 8 - 12 + 4 - 1 $$ $$ P(2) = -4 + 4 - 1 $$ $$ P(2) = 0 - 1 $$ $$ P(2) = -1 $$Answer: The remainder when $P(x) = x^3 - 3x^2 + 2x - 1$ is divided by $(x - 2)$ is $\textbf{-1}$.
Verification (Optional - using Long Division):
Let's perform the long division to confirm the remainder:
$$ \begin{array}{r} x^2 - x \phantom{+0} \\ x-2{\overline{\smash{\big)}\,x^3-3x^2+2x-1\phantom{)}}} \\ \underline{-~\phantom{(}(x^3-2x^2)\phantom{+2x-1)}} \\ -x^2+2x\phantom{-1)} \\ \underline{-~\phantom{()}(-x^2+2x)\phantom{-1)}} \\ 0-1\phantom{)} \end{array} $$The remainder is $-1$. This matches the result obtained using the Remainder Theorem.
Example 2. Find the remainder when $P(x) = 2x^4 - 5x^2 + 3$ is divided by $(x + 1)$.
Answer:
The polynomial is $P(x) = 2x^4 - 5x^2 + 3$. Note that the coefficient of $x^3$ and $x$ is $0$.
The divisor is $(x + 1)$. To use the Remainder Theorem, we need the divisor in the form $(x - c)$. We can write $(x + 1)$ as $(x - (-1))$. So, the value of $c$ is $-1$.
According to the Remainder Theorem, the remainder when $P(x)$ is divided by $(x + 1)$ is equal to $P(c)$, which is $P(-1)$.
We need to evaluate $P(x)$ at $x = -1$. Substitute $x = -1$ into the expression for $P(x)$:
$$ P(-1) = 2(-1)^4 - 5(-1)^2 + 3 $$Now, calculate the value following the order of operations:
Exponents first:
$$ (-1)^4 = (-1) \times (-1) \times (-1) \times (-1) = 1 $$ $$ (-1)^2 = (-1) \times (-1) = 1 $$Substitute these values back:
$$ P(-1) = 2(1) - 5(1) + 3 $$Multiplication next:
$$ P(-1) = 2 - 5 + 3 $$Finally, addition and subtraction from left to right:
$$ P(-1) = -3 + 3 $$ $$ P(-1) = 0 $$Answer: The remainder when $P(x) = 2x^4 - 5x^2 + 3$ is divided by $(x + 1)$ is $\textbf{0}$.
A remainder of $0$ is particularly significant. It indicates that the divisor $(x+1)$ is a factor of the polynomial $P(x)$. This leads directly to the Factor Theorem, discussed in the next section.
Applications and Significance of the Remainder Theorem
- Efficient Remainder Calculation: The most immediate use is quickly finding the remainder of polynomial division by a linear factor $(x-c)$ without performing the long division, which is much faster for higher-degree polynomials.
- Basis for the Factor Theorem: The Remainder Theorem is the direct foundation for the Factor Theorem. The Factor Theorem is essentially a special case of the Remainder Theorem where the remainder is zero.
- Polynomial Evaluation: The process used in the Remainder Theorem (substituting a value into a polynomial) is precisely how we evaluate a polynomial at a specific point. Thus, the theorem reinforces the connection between division and evaluation.
- Finding Factors and Roots: While the Remainder Theorem itself finds the remainder, if the remainder turns out to be zero, it tells us that $(x-c)$ is a factor of $P(x)$, and therefore $c$ is a root (zero) of $P(x)$. This is a key method for finding rational roots of polynomials.
- Extension to $ax+b$: The Remainder Theorem can be adapted for a divisor of the form $(ax+b)$ where $a \neq 0$. We can write $ax+b = a(x + b/a) = a(x - (-b/a))$. Dividing $P(x)$ by $ax+b$ is the same as dividing $P(x)$ by $a(x - (-b/a))$. Using the Division Algorithm, $P(x) = (ax+b) Q'(x) + R'(x)$, where degree of $R'(x)$ is less than degree of $ax+b$ (which is 1), so $R'(x)$ is a constant. Also, $P(x) = a(x - (-b/a)) Q'(x) + R'(x)$. Let $Q(x) = a Q'(x)$. Then $P(x) = (x - (-b/a)) Q(x) + R'(x)$. By the Remainder Theorem, the remainder $R'(x)$ when $P(x)$ is divided by $(x - (-b/a))$ is $P(-b/a)$. So, the remainder when $P(x)$ is divided by $(ax+b)$ is $P(-b/a)$. The Remainder Theorem applies by considering the zero of the linear divisor $ax+b$, which is $x = -b/a$.
In summary, the Remainder Theorem provides a simple and efficient way to find the remainder of polynomial division by a linear term and serves as a crucial link to the concept of polynomial roots and factors.
Factor Theorem
Connecting Zeros and Factors
The Factor Theorem is a fundamental theorem in algebra that provides a powerful link between the zeros (or roots) of a polynomial and its linear factors. It is a direct consequence of the Remainder Theorem and is often used to find factors of polynomials and, consequently, their zeros.
Essentially, the theorem states that there is a one-to-one correspondence between the zeros of a polynomial and its linear factors of the form $(x-c)$.
Statement of the Factor Theorem
Statement: Let $P(x)$ be a polynomial of degree greater than or equal to one. For any real number $c$, $(x - c)$ is a factor of $P(x)$ if and only if $P(c) = 0$.
The phrase "if and only if" (often abbreviated as "iff") means that the two parts of the statement are logically equivalent. This means the theorem consists of two implications:
- If $(x - c)$ is a factor of $P(x)$, then $P(c) = 0$. (If a linear expression divides the polynomial evenly, then evaluating the polynomial at the zero of that linear expression gives zero).
- If $P(c) = 0$, then $(x - c)$ is a factor of $P(x)$. (If evaluating the polynomial at a number $c$ gives zero, then the linear expression $(x-c)$ is a factor of the polynomial).
Proof of the Factor Theorem
The proof of the Factor Theorem relies directly on the statement of the Remainder Theorem and the definition of a factor.
Recall the Remainder Theorem: When a polynomial $P(x)$ is divided by a linear polynomial $(x - c)$, the remainder is $P(c)$. This can be formally stated using the Division Algorithm:
$$ P(x) = (x - c) \cdot Q(x) + R(x) $$where $Q(x)$ is the quotient and $R(x)$ is the remainder. Since the divisor $(x-c)$ has degree 1, the remainder $R(x)$ must have a degree less than 1, meaning $R(x)$ is a constant. By the Remainder Theorem, this constant remainder is equal to $P(c)$. So, we can write:
$P(x) = (x - c) Q(x) + P(c)$
... (i) [From Remainder Theorem]
Now let's prove the two parts of the Factor Theorem:
Proof of Part 1: If $(x - c)$ is a factor of $P(x)$, then $P(c) = 0$.
Given that $(x - c)$ is a factor of $P(x)$. By the definition of factors in polynomial division, this means that when $P(x)$ is divided by $(x-c)$, the remainder is $0$.
$\text{Remainder} = 0$
From the Remainder Theorem (or equation (i)), we know that the remainder when $P(x)$ is divided by $(x-c)$ is $P(c)$.
$\text{Remainder} = P(c)$
Equating the two expressions for the remainder, we get:
$$ P(c) = 0 $$Thus, if $(x - c)$ is a factor of $P(x)$, then $P(c) = 0$. This proves the first part of the theorem.
Proof of Part 2: If $P(c) = 0$, then $(x - c)$ is a factor of $P(x)$.
Given that $P(c) = 0$.
Consider again the result from the Remainder Theorem (equation (i)):
$$ P(x) = (x - c) Q(x) + P(c) $$Substitute the given condition $P(c) = 0$ into this equation:
$$ P(x) = (x - c) Q(x) + 0 $$ $$ P(x) = (x - c) Q(x) $$This equation shows that the polynomial $P(x)$ can be written as the product of the linear expression $(x - c)$ and some other polynomial $Q(x)$ (which is the quotient from the division). By the definition of a factor, if a polynomial can be expressed as a product of two other polynomials, then each of those polynomials is a factor of the original polynomial.
Therefore, $(x - c)$ is a factor of $P(x)$. This proves the second part of the theorem.
Since both directions of the "if and only if" statement have been proven, the Factor Theorem is established.
Examples Using the Factor Theorem
The Factor Theorem is a very useful tool for testing potential linear factors and finding zeros of polynomials.
Example 1. Show that $(x - 2)$ is a factor of the polynomial $P(x) = x^3 - 4x^2 + 5x - 2$.
Answer:
According to the Factor Theorem, $(x - 2)$ is a factor of $P(x)$ if and only if $P(2) = 0$.
The potential factor is $(x - 2)$. This matches the form $(x - c)$, where $c = 2$.
We need to evaluate the polynomial $P(x)$ at $x = 2$. Substitute $x = 2$ into $P(x)$:
$$ P(2) = (2)^3 - 4(2)^2 + 5(2) - 2 $$Calculate the value following the order of operations:
$$ P(2) = 8 - 4(4) + 10 - 2 $$ $$ P(2) = 8 - 16 + 10 - 2 $$Perform addition and subtraction from left to right:
$$ P(2) = (8 - 16) + 10 - 2 $$ $$ P(2) = -8 + 10 - 2 $$ $$ P(2) = (-8 + 10) - 2 $$ $$ P(2) = 2 - 2 $$ $$ P(2) = 0 $$Since $P(2) = 0$, by the Factor Theorem, $(x - 2)$ is indeed a factor of the polynomial $P(x) = x^3 - 4x^2 + 5x - 2$.
Example 2. Is $(x + 1)$ a factor of the polynomial $P(x) = x^4 - 1$?
Answer:
According to the Factor Theorem, $(x + 1)$ is a factor of $P(x)$ if and only if $P(-1) = 0$.
The potential factor is $(x + 1)$. We can write this in the form $(x - c)$ as $(x - (-1))$. So, the value of $c$ is $-1$.
We need to evaluate the polynomial $P(x)$ at $x = -1$. Substitute $x = -1$ into $P(x)$:
$$ P(-1) = (-1)^4 - 1 $$Calculate the value:
Exponent first: $(-1)^4 = (-1) \times (-1) \times (-1) \times (-1) = 1$.
$$ P(-1) = 1 - 1 $$ $$ P(-1) = 0 $$Since $P(-1) = 0$, by the Factor Theorem, $(x + 1)$ is a factor of the polynomial $P(x) = x^4 - 1$.
Verification (Optional - using Factoring):
We know that $x^4 - 1$ is a difference of squares: $(x^2)^2 - 1^2$.
$$ x^4 - 1 = (x^2 - 1)(x^2 + 1) $$$x^2 - 1$ is also a difference of squares:
$$ x^2 - 1 = (x - 1)(x + 1) $$So, $P(x) = (x - 1)(x + 1)(x^2 + 1)$. Since $(x+1)$ appears as a factor in the factored form, our conclusion from the Factor Theorem is correct.
Example 3. Find the value of $k$ if $(x - 1)$ is a factor of the polynomial $P(x) = 2x^2 + kx - 3$.
Answer:
We are given that $(x - 1)$ is a factor of $P(x) = 2x^2 + kx - 3$.
According to the Factor Theorem, if $(x - 1)$ is a factor of $P(x)$, then $P(1)$ must be equal to $0$.
The potential factor is $(x - 1)$. This matches the form $(x - c)$, where $c = 1$.
We need to evaluate the polynomial $P(x)$ at $x = 1$ and set the result equal to $0$. Substitute $x = 1$ into $P(x)$:
$$ P(1) = 2(1)^2 + k(1) - 3 $$Set this equal to $0$:
$$ 2(1)^2 + k(1) - 3 = 0 $$Simplify the equation:
$$ 2(1) + k - 3 = 0 $$ $$ 2 + k - 3 = 0 $$Combine the constant terms:
$$ k - 1 = 0 $$Solve for $k$:
$$ k = 1 $$Answer: The value of $k$ for which $(x - 1)$ is a factor of $P(x)$ is $\textbf{1}$. The polynomial is $P(x) = 2x^2 + x - 3$.
Verification (Optional):
Substitute $k=1$ back into the polynomial: $P(x) = 2x^2 + x - 3$.
Now test if $(x-1)$ is a factor by evaluating $P(1)$:
$$ P(1) = 2(1)^2 + (1) - 3 = 2(1) + 1 - 3 = 2 + 1 - 3 = 3 - 3 = 0 $$Since $P(1)=0$, the Factor Theorem confirms that $(x-1)$ is a factor when $k=1$.
Applications and Significance of the Factor Theorem
- Testing for Factors: It provides a quick and easy way to check if a given linear expression of the form $(x-c)$ is a factor of a polynomial by simply evaluating the polynomial at $x=c$.
- Finding Zeros: If you find a value $c$ such that $P(c)=0$ (perhaps by trial and error with rational root possibilities, which is a more advanced technique), the Factor Theorem immediately tells you that $c$ is a zero of the polynomial, and $(x-c)$ is a factor.
- Factoring Polynomials: Once you find a linear factor $(x-c)$ using the Factor Theorem, you can use polynomial long division (or synthetic division) to divide $P(x)$ by $(x-c)$ and find the quotient $Q(x)$. Since $P(x) = (x-c)Q(x)$, you have partially factored the polynomial. You can then try to factor $Q(x)$ further to find more linear factors and all the zeros. This is a standard approach for factoring and finding zeros of polynomials of degree 3 or higher.
- Constructing Polynomials: If you know the zeros of a polynomial, say $\alpha_1, \alpha_2, \ldots, \alpha_n$, the Factor Theorem allows you to write the polynomial in its factored form: $P(x) = a_n (x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_n)$, where $a_n$ is the leading coefficient.
The Factor Theorem is a cornerstone concept that connects the analytical process of finding polynomial roots with the algebraic structure of polynomial factorisation.
Division of Polynomial by Polynomial
Polynomial Long Division
When the divisor in polynomial division is not a monomial (i.e., it is a polynomial with two or more terms, such as a binomial or trinomial), we use a process called polynomial long division. This method is very similar to the long division process you use for dividing integers and is a systematic way to find the quotient and remainder polynomials, as guaranteed by the Division Algorithm for Polynomials.
Polynomial long division is a key technique for:
- Simplifying rational expressions where the numerator's degree is greater than or equal to the denominator's degree.
- Finding factors of polynomials.
- Determining the roots (zeros) of polynomial equations, especially when combined with the Factor Theorem.
The Process of Polynomial Long Division
To divide a polynomial $P(x)$ (the dividend) by a non-zero polynomial $G(x)$ (the divisor), where the degree of $P(x)$ is greater than or equal to the degree of $G(x)$, follow these steps:
Set Up the Division:
Write the division problem in the traditional long division format. Both the dividend $P(x)$ and the divisor $G(x)$ must be written in standard form, which means arranging the terms in decreasing order of their degrees (e.g., $x^3$ then $x^2$ then $x$ then constant). If the dividend is missing any terms for powers of the variable between the highest and lowest degree terms, include those terms with a coefficient of $0$. For example, if the dividend is $x^3 + 5x - 2$, write it as $x^3 + 0x^2 + 5x - 2$. This helps in aligning terms during subtraction.Divide Leading Terms:
Divide the leading term (the term with the highest degree) of the current dividend by the leading term of the divisor. This result is the first term of your quotient.Multiply:
Multiply the term you just found in the quotient (from Step 2) by the entire divisor polynomial $G(x)$.Subtract:
Subtract the polynomial obtained in Step 3 from the current dividend. It is crucial to be careful with the signs during this subtraction; subtracting a polynomial is equivalent to adding its additive inverse (changing the sign of each term and then adding). This subtraction will eliminate the leading term of the current dividend. The result of this subtraction is a new polynomial.Bring Down:
Bring down the next term (or terms) from the original dividend to the new polynomial obtained in Step 4. Bring down enough terms so that the degree of the resulting polynomial is greater than or equal to the degree of the divisor, allowing you to continue the division process.Repeat:
Treat the polynomial resulting from Step 5 as the new dividend. Repeat the entire process from Step 2: Divide the leading term of this new dividend by the leading term of the original divisor to find the next term of the quotient. Multiply this new quotient term by the divisor, subtract from the new dividend, and bring down terms as needed.Terminate:
Continue this iterative process until the degree of the polynomial resulting from the subtraction is strictly less than the degree of the divisor $G(x)$, or until the remainder is $0$. The polynomial left at this point is the remainder $R(x)$, and the polynomial accumulated at the top is the quotient $Q(x)$.
Examples of Polynomial Long Division
Let's apply the steps to specific examples.
Example 1. Divide $P(x) = x^2 - 5x + 6$ by $G(x) = x - 2$.
Answer:
Set up the long division. Both polynomials are already in standard form and have no missing terms.
$$ \begin{array}{r} \phantom{x} x - 3 \\ x-2{\overline{\smash{\big)}\,x^2-5x+6}} \\ \end{array} $$Step 1: Divide the leading term of the dividend ($x^2$) by the leading term of the divisor ($x$).
$$ \frac{x^2}{x} = x $$This is the first term of the quotient. Write it above the $x$ term in the dividend:
$$ \begin{array}{r} x \phantom{-3} \\ x-2{\overline{\smash{\big)}\,x^2-5x+6}} \\ \end{array} $$Step 2: Multiply the first quotient term ($x$) by the entire divisor ($x-2$).
$$ x(x - 2) = x^2 - 2x $$Write this result below the dividend, aligning like terms:
$$ \begin{array}{r} x \phantom{-3} \\ x-2{\overline{\smash{\big)}\,x^2-5x+6}} \\ x^2 - 2x \phantom{+6} \\ \hline \end{array} $$Step 3: Subtract the result from the dividend. Change the signs of the terms being subtracted and add:
$$ (x^2 - 5x) - (x^2 - 2x) = x^2 - 5x - x^2 + 2x = -3x $$ $$ \begin{array}{r} x \phantom{-3} \\ x-2{\overline{\smash{\big)}\,x^2-5x+6}} \\ \underline{-~(x^2 - 2x)} \\ \phantom{x^2} -3x \phantom{+6} \\ \end{array} $$Step 4: Bring down the next term from the dividend ($+6$). The new polynomial to work with is $-3x + 6$.
$$ \begin{array}{r} x \phantom{-3} \\ x-2{\overline{\smash{\big)}\,x^2-5x+6}} \\ \underline{-~(x^2 - 2x)} \\ \phantom{x^2} -3x + 6 \\ \end{array} $$Step 5: Now, treat $-3x + 6$ as the new dividend. Divide its leading term ($-3x$) by the leading term of the divisor ($x$).
$$ \frac{-3x}{x} = -3 $$This is the next term of the quotient. Write it above the constant term in the dividend:
$$ \begin{array}{r} x - 3 \\ x-2{\overline{\smash{\big)}\,x^2-5x+6}} \\ \underline{-~(x^2 - 2x)} \\ \phantom{x^2} -3x + 6 \\ \end{array} $$Step 6: Multiply the new quotient term ($-3$) by the entire divisor ($x-2$).
$$ -3(x - 2) = -3x + 6 $$Write this result below the current dividend ($-3x + 6$), aligning like terms:
$$ \begin{array}{r} x - 3 \\ x-2{\overline{\smash{\big)}\,x^2-5x+6}} \\ \underline{-~(x^2 - 2x)} \\ \phantom{x^2} -3x + 6 \\ -3x + 6 \\ \hline \end{array} $$Step 7: Subtract the result from the current dividend. Change the signs of the terms being subtracted and add:
$$ (-3x + 6) - (-3x + 6) = -3x + 6 + 3x - 6 = 0 $$ $$ \begin{array}{r} x - 3 \\ x-2{\overline{\smash{\big)}\,x^2-5x+6}} \\ \underline{-~(x^2 - 2x)} \\ \phantom{x^2} -3x + 6 \\ \underline{-~(-3x + 6)} \\ \phantom{-3x+6} 0 \\ \end{array} $$The resulting polynomial is $0$. The degree of $0$ is undefined (or $-\infty$), which is less than the degree of the divisor ($x-2$, degree 1). The division process terminates.
The polynomial above the division symbol is the quotient: $Q(x) = \textbf{x - 3}$.
The final result at the bottom is the remainder: $R(x) = \textbf{0}$.
Check with the Division Algorithm: $P(x) = G(x)Q(x) + R(x)$.
$$ x^2 - 5x + 6 = (x - 2)(x - 3) + 0 $$Expand the right side: $(x - 2)(x - 3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6$. This matches the dividend. The result is correct.
Example 2. Divide $P(x) = 2x^3 + 3x^2 - 4x + 5$ by $G(x) = x + 2$.
Answer:
Set up the long division. Both polynomials are in standard form and have no missing terms.
$$ \begin{array}{r} \phantom{2x^2} \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ \end{array} $$Step 1: Divide the leading term of the dividend ($2x^3$) by the leading term of the divisor ($x$).
$$ \frac{2x^3}{x} = 2x^2 $$First term of quotient: $2x^2$. Write it above the $x^2$ term.
$$ \begin{array}{r} 2x^2 \phantom{-x-2} \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ \end{array} $$Step 2: Multiply $2x^2$ by the divisor ($x+2$).
$$ 2x^2(x + 2) = 2x^3 + 4x^2 $$Write result below dividend:
$$ \begin{array}{r} 2x^2 \phantom{-x-2} \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ 2x^3 + 4x^2 \phantom{-4x+5} \\ \hline \end{array} $$Step 3: Subtract $(2x^3 + 4x^2)$ from $(2x^3 + 3x^2 - 4x + 5)$. Change signs and add:
$$ (2x^3 + 3x^2) - (2x^3 + 4x^2) = 2x^3 + 3x^2 - 2x^3 - 4x^2 = -x^2 $$ $$ \begin{array}{r} 2x^2 \phantom{-x-2} \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ \underline{-~(2x^3 + 4x^2)} \\ \phantom{2x^3} -x^2 \phantom{-4x+5} \\ \end{array} $$Step 4: Bring down the next term from the dividend ($-4x$). New polynomial: $-x^2 - 4x$.
$$ \begin{array}{r} 2x^2 \phantom{-x-2} \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ \underline{-~(2x^3 + 4x^2)} \\ \phantom{2x^3} -x^2 - 4x \phantom{+5} \\ \end{array} $$Step 5: Treat $-x^2 - 4x$ as the new dividend. Divide its leading term ($-x^2$) by the leading term of the divisor ($x$).
$$ \frac{-x^2}{x} = -x $$Next term of quotient: $-x$. Write it above the $x$ term.
$$ \begin{array}{r} 2x^2 - x \phantom{-2} \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ \underline{-~(2x^3 + 4x^2)} \\ \phantom{2x^3} -x^2 - 4x \phantom{+5} \\ \end{array} $$Step 6: Multiply $-x$ by the divisor ($x+2$).
$$ -x(x + 2) = -x^2 - 2x $$Write result below current dividend:
$$ \begin{array}{r} 2x^2 - x \phantom{-2} \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ \underline{-~(2x^3 + 4x^2)} \\ \phantom{2x^3} -x^2 - 4x \phantom{+5} \\ -x^2 - 2x \phantom{+5} \\ \hline \end{array} $$Step 7: Subtract $(-x^2 - 2x)$ from $(-x^2 - 4x)$. Change signs and add:
$$ (-x^2 - 4x) - (-x^2 - 2x) = -x^2 - 4x + x^2 + 2x = -2x $$ $$ \begin{array}{r} 2x^2 - x \phantom{-2} \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ \underline{-~(2x^3 + 4x^2)} \\ \phantom{2x^3} -x^2 - 4x \phantom{+5} \\ \underline{-~(-x^2 - 2x)} \\ \phantom{-x^2} -2x \phantom{+5} \\ \end{array} $$Step 8: Bring down the next term from the dividend ($+5$). New polynomial: $-2x + 5$.
$$ \begin{array}{r} 2x^2 - x \phantom{-2} \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ \underline{-~(2x^3 + 4x^2)} \\ \phantom{2x^3} -x^2 - 4x \phantom{+5} \\ \underline{-~(-x^2 - 2x)} \\ \phantom{-x^2} -2x + 5 \\ \end{array} $$Step 9: Treat $-2x + 5$ as the new dividend. Divide its leading term ($-2x$) by the leading term of the divisor ($x$).
$$ \frac{-2x}{x} = -2 $$Next term of quotient: $-2$. Write it above the constant term.
$$ \begin{array}{r} 2x^2 - x - 2 \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ \underline{-~(2x^3 + 4x^2)} \\ \phantom{2x^3} -x^2 - 4x \phantom{+5} \\ \underline{-~(-x^2 - 2x)} \\ \phantom{-x^2} -2x + 5 \\ \end{array} $$Step 10: Multiply $-2$ by the divisor ($x+2$).
$$ -2(x + 2) = -2x - 4 $$Write result below current dividend:
$$ \begin{array}{r} 2x^2 - x - 2 \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ \underline{-~(2x^3 + 4x^2)} \\ \phantom{2x^3} -x^2 - 4x \phantom{+5} \\ \underline{-~(-x^2 - 2x)} \\ \phantom{-x^2} -2x + 5 \\ -2x - 4 \\ \hline \end{array} $$Step 11: Subtract $(-2x - 4)$ from $(-2x + 5)$. Change signs and add:
$$ (-2x + 5) - (-2x - 4) = -2x + 5 + 2x + 4 = 9 $$ $$ \begin{array}{r} 2x^2 - x - 2 \\ x+2{\overline{\smash{\big)}\,2x^3+3x^2-4x+5}} \\ \underline{-~(2x^3 + 4x^2)} \\ \phantom{2x^3} -x^2 - 4x \phantom{+5} \\ \underline{-~(-x^2 - 2x)} \\ \phantom{-x^2} -2x + 5 \\ \underline{-~(-2x - 4)} \\ \phantom{-2x+5} 9 \\ \end{array} $$The resulting polynomial is $9$. This is a constant polynomial with degree $0$. The degree of the divisor ($x+2$) is $1$. Since $0 < 1$, the degree of the remainder is less than the degree of the divisor. The division process terminates.
The quotient is $Q(x) = \textbf{2x\textsuperscript{2} - x - 2}$.
The remainder is $R(x) = \textbf{9}$.
Check with the Division Algorithm: $P(x) = G(x)Q(x) + R(x)$.
$$ 2x^3 + 3x^2 - 4x + 5 = (x + 2)(2x^2 - x - 2) + 9 $$Expand the product $(x + 2)(2x^2 - x - 2)$:
$$ x(2x^2 - x - 2) + 2(2x^2 - x - 2) $$ $$ (x \times 2x^2 - x \times x - x \times 2) + (2 \times 2x^2 - 2 \times x - 2 \times 2) $$ $$ (2x^3 - x^2 - 2x) + (4x^2 - 2x - 4) $$Combine like terms:
$$ 2x^3 + (-x^2 + 4x^2) + (-2x - 2x) - 4 $$ $$ 2x^3 + 3x^2 - 4x - 4 $$Now, add the remainder $R(x) = 9$ to this product:
$$ (2x^3 + 3x^2 - 4x - 4) + 9 $$ $$ 2x^3 + 3x^2 - 4x + (-4 + 9) $$ $$ 2x^3 + 3x^2 - 4x + 5 $$This matches the original dividend $P(x)$. The result is correct.
Polynomial long division is a methodical process. Practicing several examples helps in becoming proficient with the steps, particularly managing the subtraction and aligning terms correctly. It is a fundamental skill for further topics in polynomial algebra.